# Chisquare Test

Pearson's chi-squared test is a statistical test applied to sets of categorical data to evaluate how likely it is that any observed difference between the sets arose by chance. It is the most widely used of many chi-squared tests (e.g., Yates, likelihood ratio, portmanteau test in time series, etc.) – Statistical procedures whose results are evaluated by reference to the chi-squared distribution.

It tests a null hypothesis stating that the frequency distribution of certain events observed in a sample is consistent with a particular theoretical distribution. The events considered must be mutually exclusive and have total probability 1. A common case for this is where the events each cover an outcome of a categorical variable. A simple example is the hypothesis that an ordinary six-sided die is "fair" (i. e., all six outcomes are equally likely to occur).

## Definition

Pearson's chi-squared test is used to assess two types of comparison: tests of goodness of fit and tests of independence.

• A test of goodness of fit establishes whether or not an observed frequency distribution differs from a theoretical distribution.
• A test of independence assesses whether unpaired observations on two variables, expressed in a contingency table, are independent of each other (e.g. polling responses from people of different nationalities to see if one's nationality is related to the response).

The procedure of the test includes the following steps:

Chi-squared Test
1. Calculate the chi-squared test statistic, $\chi^2$, which resembles a normalized sum of squared deviations between observed and theoretical frequencies (see below).
2. Determine the degrees of freedom, df, of that statistic, which is essentially the number of categories minus the number of parameters of the fitted distribution.
3. Select a desired level of confidence (significance level, p-value or alpha level) for the result of the test.
4. Compare $\chi^2$ to the critical value from the chi-squared distribution with df degrees of freedom and the selected confidence level (one-sided since the test is only one direction, i.e. is the test value greater than the critical value?), which in many cases gives a good approximation of the distribution of $\chi^2$.
5. Accept or reject the null hypothesis that the observed frequency distribution is different from the theoretical distribution based on whether the test statistic exceeds the critical value of $\chi^2$. If the test statistic exceeds the critical value of $\chi^2$, the null hypothesis ($H_0$ = there is no difference between the distributions) can be rejected with the selected level of confidence and the alternative hypothesis ($H_1$ = there is a difference between the distributions) can be accepted with the selected level of confidence.

## Test for fit of a distribution

### Discrete uniform distribution

In this case $N$ observations are divided among $n$ cells. A simple application is to test the hypothesis that, in the general population, values would occur in each cell with equal frequency. The "theoretical frequency" for any cell (under the null hypothesis of a discrete uniform distribution) is thus calculated as $E_i=\frac{N}{n}$ and the reduction in the degrees of freedom is $p=1$, notionally because the observed frequencies $O_i$ are constrained to sum to $N$.

### Other distributions

When testing whether observations are random variables whose distribution belongs to a given family of distributions, the "theoretical frequencies" are calculated using a distribution from that family fitted in some standard way. The reduction in the degrees of freedom is calculated as $p=s+1$, where $s$ is the number of parameters used in fitting the distribution. For instance, when checking a normal distribution (where the parameters are mean and standard deviation), $p=3$, and when checking a Poisson distribution (where the parameter is the expected value), $p=2$. Thus, there will be $n-p$ degrees of freedom, where $n$ is the number of categories.

It should be noted that the degrees of freedom are not based on the number of observations as with a Student's t or F-distribution. For example, if testing for a fair, six-sided die, there would be five degrees of freedom because there are six categories/parameters (each number). The number of times the die is rolled does not influence the number of degrees of freedom.

### Calculating the test-statistic

The value of the test-statistic is

[$] \chi^2 = \sum_{i=1}^{n} \frac{(O_i - E_i)^2}{E_i} = N \sum_{i=1}^n \frac{\left(O_i/N - p_i \right)^2}{p_i} [$]

where

$\chi^2$ = Pearson's cumulative test statistic, which asymptotically approaches a $\chi^2$ distribution.
$O_i$ = the number of observations of type i.
$N$ = total number of observations
$E_i$ = $N p_i$ = the expected (theoretical) frequency of type $i$, asserted by the null hypothesis that the fraction of type $i$ in the population is $p_i$
$n$ = the number of cells in the table.

The chi-squared statistic can then be used to calculate a p-value by comparing the value of the statistic to a chi-squared distribution. The number of degrees of freedom is equal to the number of cells $n$, minus the reduction in degrees of freedom, $p$.

## Test of independence

In this case, an "observation" consists of the values of two outcomes and the null hypothesis is that the occurrence of these outcomes is statistically independent. Each observation is allocated to one cell of a two-dimensional array of cells (called a contingency table) according to the values of the two outcomes. If there are $r$ rows and $c$ columns in the table, the "theoretical frequency" for a cell, given the hypothesis of independence, is

[$]E_{i,j}= N p_{i\cdot} p_{\cdot j} ,[$]

where $N$ is the total sample size (the sum of all cells in the table), and

[$] p_{i\cdot} = \frac{O_{i\cdot}}{N} = \sum_{j=1}^c \frac{O_{i,j}}{N},[$]

is the fraction of observations of type $i$ ignoring the column attribute (fraction of row totals), and

[$] p_{\cdot j} = \frac{O_{\cdot j}}{N} = \frac{\sum_{i = 1}^r O_{i,j}}{N} [$]

is the fraction of observations of type $j$ ignoring the row attribute (fraction of column totals). The term frequencies" refers to absolute numbers rather than already normalised values.

The value of the test-statistic is

[$] \chi^2 = \sum_{i=1}^{r} \sum_{j=1}^{c} {(O_{i,j} - E_{i,j})^2 \over E_{i,j}} = N \sum_{i,j} p_{i\cdot}p_{\cdot j} \left (\frac{(O_{i,j}/N) - p_{i\cdot}p_{\cdot j}}{p_{i\cdot}p_{\cdot j}} \right )^2 [$]

Note that $\chi^2$ is 0 if and only if $O_{i,j} = E_{i,j}$ for all $i,j$ -- only if the expected and true number of observations are equal in all cells.

Fitting the model of "independence" reduces the number of degrees of freedom by $p = r + 1 - c$. The number of degrees of freedom is equal to the number of cells $rc$, minus the reduction in degrees of freedom, $p$, which reduces to $(r-1)(c-1)$.

For the test of independence, also known as the test of homogeneity, a chi-squared probability of less than or equal to 0.05 (or the chi-squared statistic being at or larger than the 0.05 critical point) is commonly interpreted by practicionners as justification for rejecting the null hypothesis that the row variable is independent of the column variable.[1] The alternative hypothesis corresponds to the variables having an association or relationship where the structure of this relationship is not specified.

## Assumptions

The chi-squared test, when used with the standard approximation that a chi-squared distribution is applicable, has the following assumptions:

Assumption Description
Simple random sample The sample data is a random sampling from a fixed distribution or population where every collection of members of the population of the given sample size has an equal probability of selection.
Sample size (whole table) A sample with a sufficiently large size is assumed. If a chi squared test is conducted on a sample with a smaller size, then the chi squared test will yield an inaccurate inference. The researcher, by using chi squared test on small samples, might end up committing a Type II error.
Expected cell count Adequate expected cell counts. Some require 5 or more, and others require 10 or more. A common rule is 5 or more in all cells of a 2-by-2 table, and 5 or more in 80% of cells in larger tables, but no cells with zero expected count. When this assumption is not met, Yates's correction is applied.
Independence The observations are always assumed to be independent of each other. This means chi-squared cannot be used to test correlated data (like matched pairs or panel data).

## Limitations

The approximation to the chi-squared distribution breaks down if expected frequencies are too low. It will normally be acceptable so long as no more than 20% of the events have expected frequencies below 5. Where there is only 1 degree of freedom, the approximation is not reliable if expected frequencies are below 10. In this case, a better approximation can be obtained by reducing the absolute value of each difference between observed and expected frequencies by 0.5 before squaring; this is called Yates's correction for continuity.

## References

1. "Critical Values of the Chi-Squared Distribution". NIST/SEMATECH e-Handbook of Statistical Methods. National Institute of Standards and Technology.